1. Use freezing point depression formula

For association/dissociation, $\Delta T_f = i K_f m$ where:

• $\Delta T_f = 0.2^\circ C$
• $K_f = 1.86\, \text{K kg mol}^{-1}$
• $i$ = van't Hoff factor
• $m$ = molality

2. Calculate molality assuming no association

Moles of solute A: $n = \frac{0.7}{93} = 0.00753\, \text{mol}$

Mass of water: $42.0\, \text{g} = 0.042\, \text{kg}$

Molality: $m = \frac{0.00753}{0.042} = 0.1793\, \text{mol kg}^{-1}$

3. Find van't Hoff factor $i$

$i = \frac{\Delta T_f}{K_f m} = \frac{0.2}{1.86 \times 0.1793}$

$i \approx \frac{0.2}{0.3335} \approx 0.60$

4. Relate $i$ to association

Since $i<1$, solute associates. Assume dimerization: $2A \rightarrow A_2$

If degree of association is $\alpha$, then for dimerization: $i = 1 - \frac{\alpha}{2}$

Substitute $i = 0.60$: $0.60 = 1 - \frac{\alpha}{2}$

$\frac{\alpha}{2} = 0.40$

$\alpha = 0.80$

Thus percentage association is $0.80 \times 100 = 80\%$

5. Check options

• A: $50\%$ ✗
• B: $60\%$ ✗
• C: $70\%$ ✗
• D: $80\%$ ✓

Therefore, the correct option is D.